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Let $r' = r - a$

I am trying to proof that neighbourhoods or open balls is an open set. I've watched this video on youtube where the guy explains how the triangle inequality comes into play. I quote

Anything that's distance less than r' from q is going to be distance less than r from p

Can someone show me how that is done? All I can write is that for some x that's between the q and r', and p and r. I get $|x - q| < r' < |x - p| < r$

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jak, do you mean to say let $r^\prime = r-a$? –  amWhy Nov 2 '12 at 22:05
    
@amWhy, yes I did... let me fix that –  jip Nov 2 '12 at 22:10
1  
«Neighborhood» and «open ball» do not mean the same thing. –  Mariano Suárez-Alvarez Nov 2 '12 at 22:26
    
I thought they do mean the same thing. I am reading a book by Rudin and it seems like that's what they are implying with the "radius" of the neighborhood. He even mentioned an example that in $\mathbb{R^2}$, the neighbourhoods are the interiors of the circle –  jip Nov 3 '12 at 1:02

1 Answer 1

up vote 2 down vote accepted

Let $r'=r-a$. Suppose that $|x-q|<r'$. Then $$|x-p|\le|x-q|+|q-p|<r'+a=(r-a)+a=r\;,$$

and you’re done. The first step is the triangle inequality, the next is the assumption that $|x-q|<r'$ together with the fact that $|q-p|=a$, and the rest is just algebra.

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